How to Fourier Transform $\frac{\sin(x)^2}{x^2}$ without Contour Integration. In our lecture we need to Fourier transform $\frac{\sin(x)^2}{x^2}$, i.e. compute the integral:
$$\int_{-\infty}^{\infty} \mathrm e^{-iy x}\frac{\sin(x)^2}{x^2} \mathrm dx$$ 
Since it's a lecture on partial differential equations and not complex analysis, I don't think contour integration can be the solution here. I already tried to rewrite $\sin x$ getting 
$$\int_{-\infty}^{\infty}\frac{-1}{4x^2} (\mathrm e^{ix(2-y)}-2\mathrm e^{-ixy}+\mathrm e^{-ix(y+2)})\mathrm dx$$
but now I'm still stuck with the computation of something like $$\int_{-\infty}^{\infty}\frac{\mathrm e^{-ixy}}{x^2}\mathrm dx$$
Any ideas or hints are greatly appreciated. Thanks in advance!
 A: Due to parity it is enough to compute
$$\int_{-\infty}^{+\infty}\frac{\sin^2(x)\cos(xs)}{x^2}\,dx \stackrel{\text{def}}{=}\lim_{M\to +\infty}\int_{-M}^{M}\frac{\sin^2(x)\cos(xs)}{x^2}\,dx $$
and by integration by parts the RHS equals
$$ \lim_{M\to +\infty}\int_{-M}^{M}\frac{\frac{1}{2}\cos(xs)-\frac{1}{4}\cos(x(s+2))-\frac{1}{4}\cos(x(s-2))}{x^2}\,dx $$
or
$$ \lim_{M\to +\infty}\int_{-M}^{M}\frac{\frac{s+2}{4}\sin(x(s+2))+\frac{s-2}{4}\sin(x(s-2))+\frac{s}{2}\sin(xs)}{x}\,dx $$
where we may exploit the standard result
$$ \forall \alpha\in\mathbb{R},\qquad \int_{-\infty}^{+\infty}\frac{\sin(\alpha x)}{x}\,dx = \pi\,\text{Sign}(\alpha) $$
to get:
$$\int_{-\infty}^{+\infty}\frac{\sin^2(x)\cos(xs)}{x^2}\,dx =\frac{\pi}{4}\left[|s-2|+|s+2|-2|s|\right]. $$
The RHS is a piecewise-linear function, supported on $[-2,2]$, going from $0$ to $\pi$ on $[-2,0]$ and from $\pi$ to $0$ on $[0,2]$. Not by chance, it is a multiple of the convolution between $\mathbb{1}_{(-1,1)}$ and itself.
A: This is trivial from the Inversion Theorem for the Fourier transform.
Note that below I leave out all the $\pi$'s and $2\pi$'s; to actually get the right answer you'll need to look up some definitions and insert a few constants.
Note first that $\frac{\sin(x)}x=\hat f(x)$, where $f=\frac12\chi_{[-1,1]}$.
So $\frac{\sin^2(x)}{x^2}=\widehat{f*f}$. Since all our functions are even the inversion theorem shows that the Fourier transform of $\frac{\sin^2(x)}{x^2}$ is $f*f$, which you can easily calculate. (In fact $f*f$ should be a triangle function supported on $[-2,2]$, which agrees with the other answer.)
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\ds{\int_{-\infty}^{\infty}\expo{-\ic yx}\,{\sin^{2}\pars{x} \over x^{2}}\,\dd x}}  =
\int_{-\infty}^{\infty}\expo{-\ic yx}\
\overbrace{\bracks{{1 \over 2}\int_{-1}^{1}\expo{\ic kx}\,\dd k}}
^{\ds{\sin\pars{x} \over x}}\
\overbrace{\bracks{{1 \over 2}\int_{-1}^{1}\expo{\ic qx}\,\dd q}}
^{\ds{\sin\pars{x} \over x}}\ \,\dd x
\\[5mm] = &\
{1 \over 4}\int_{-1}^{1}\int_{-1}^{1}\
\overbrace{\int_{-\infty}^{\infty}\expo{\pars{-y + k + q}x\ic}\,\dd x}
^{\ds{2\pi\,\delta\pars{-y + k + q}}}\ \dd k\,\dd q =
{1 \over 2}\,\pi\int_{-1}^{1}\int_{-1}^{1}\delta\pars{k - \bracks{y - q}}
\dd k\,\dd q
\\[5mm] = &\
{1 \over 2}\,\pi\int_{-1}^{1}\bracks{-1 < y - q < 1}\,\dd q =
{1 \over 2}\,\pi\int_{-1}^{1}\bracks{y - 1 < q < y + 1}\,\dd q
\\[5mm] = &\
{1 \over 2}\,\pi\braces{%
\bracks{y - 1 < -1}\bracks{-1 < y + 1 < 1}\int_{-1}^{y + 1}\dd q +
\bracks{-1 < y - 1 < 1}\bracks{y + 1 > 1}\int_{y - 1}^{1}\dd q}
\\[5mm] = &\
{1 \over 2}\,\pi\braces{\vphantom{\Large A}%
\bracks{y < 0}\bracks{-2 < y < 0}\pars{y + 2} +
\bracks{0 < y < 2}\bracks{y > 0}\pars{2 - y}}
\\[5mm] = &\
{1 \over 2}\,\pi\braces{\vphantom{\Large A}%
\bracks{-2 < y < 0}\pars{2 - \verts{y}} +
\bracks{0 < y < 2}\bracks{y > 0}\pars{2 - \verts{y}}}
\\[5mm] = &\
\bbx{\pars{1 - {1 \over 2}\,\verts{y}}\,\pi
\bracks{\vphantom{\large A}\verts{y} < 2}\bracks{\vphantom{\large A}y \not= 0}}
\end{align}

Note that the above result $\ds{\to \pi}$ as $\ds{y \to 0}$ !!!.


